An old puzzle about two numbers

XYYXF · 2007-02-21, 00:22

Two integers are chosen. It is known than both of them are greater than 1 and less than 100. It's also known that Mersenne was informed of their sum, while Fermat was informed of their product. So, the following conversation between them took place:

Fermat: Of course you know that I'm not able to find the numbers chosen...
Mersenne: You are right. And I can't find them too...
Fermat: But I've just found them!

What numbers were chosen, and how did Mersenne and Fermat draw their conclusions?

Of course both of them were honest

davieddy · 2007-02-21, 09:51

I assume that only one pair of numbers have a product which
can be formed in more than one way and the sum of the two numbers
is the same for more than one case. Does this make sense?
I'll try it in Basic.

David

akruppa · 2007-02-21, 10:31

Lets say the two numbers are a, b, with a<b.
The first statement means the two numbers a,b are not both prime, their product is not the cube of a prime, and for at least two k|a*b with 1<k<100, a*b/k is < 100. I.e., a*b wasn't 2*31*59, as that would only allow a=59, b=62.
The second statement means that the sum Mersenne was given can be written in more than one way as a+b with such a,b. Also, the sum isn't 198 or 197.

It may be possible to solve it from there with an exhaustive search.

Alex

Patrick123 · 2007-02-21, 13:04

The first numbers that spring to mind are 4 and 3. Giving the product 12 and the sum 7.

12 can be expressed as 4*3 or 6*2 so Fermat could not define them.
7 can be expressed as 4+3 or 5+2 likewise for Mersenne.
The statement did say the numbers where greater than 1 and other numbers would not give such a simple addition.

I have not thought this completely through but it seems to give a quick solution.

Regards
Patrick

akruppa · 2007-02-21, 13:17

But Fermat said he *knew* that Mersenne knows that Fermat can't figure out the numbers. If Fermat had been given the product 12, he would have known that Mersenne must have been given the sum 7 or 8. With the sum 7, Mersenne could not have known for certain that Fermat was stuck, as it might have admitted a=2, b=5, which Fermat could have solved...

Nice little mind twister.

Alex

akruppa · 2007-02-21, 13:25

So I guess the sum a+b was an odd number, as any even number can (conjecturally) be written as the sum of two primes, and a+b-2 must not be prime.

Alex

Patrick123 · 2007-02-21, 13:30

Ah! I see what you mean. I thought it was too easy.

Regards
Patrick

XYYXF · 2007-02-21, 14:49

akruppa is on the right way :)

Patrick123 · 2007-02-21, 22:09

The two numbers are 9 and 21. Product 189 - Sum 30.

Taking Alex's lead, I used Excel and filtered out:
All Evens.
All primes.
All products of two primes.
All numbers where n-2 was prime.

I then factorized the remaining numbers and created a permutation of all the ways you can sum the numbers. There where 2 products that had the same sum with one being a cube. This is all Fermat needed to know to solve this.

Regards
Patrick

XYYXF · 2007-02-22, 23:31

The answer (9, 21) is wrong :)

rogue · 2007-02-23, 00:11

I saw this puzzle on the net and the following line is included:

Both are unequal to 1 and the sum of them is less than 100.

This is different than XYYXF's statement. I don't know if this is supposed to be part of this puzzle or not.

2007-02-21, 00:22	#1
XYYXF Jan 2005 Minsk, Belarus 110010000₂ Posts	An old puzzle about two numbers Two integers are chosen. It is known than both of them are greater than 1 and less than 100. It's also known that Mersenne was informed of their sum, while Fermat was informed of their product. So, the following conversation between them took place: Fermat: Of course you know that I'm not able to find the numbers chosen... Mersenne: You are right. And I can't find them too... Fermat: But I've just found them! What numbers were chosen, and how did Mersenne and Fermat draw their conclusions? Of course both of them were honest Last fiddled with by XYYXF on 2007-02-21 at 00:23

2007-02-21, 10:31	#3
akruppa "Nancy" Aug 2002 Alexandria 2,467 Posts	Lets say the two numbers are a, b, with a<b. The first statement means the two numbers a,b are not both prime, their product is not the cube of a prime, and for at least two k\|ab with 1<k<100, ab/k is < 100. I.e., ab wasn't 23159, as that would only allow a=59, b=62. The second statement means that the sum Mersenne was given can be written in more than one way as a+b with such a,b. Also, the sum isn't 198 or 197. It may be possible to solve it from there with an exhaustive search. Alex Last fiddled with by akruppa on 2007-02-21 at 22:33 Reason: 62, not 61*

2007-02-21, 13:04	#4
Patrick123 Jan 2006 JHB, South Africa 157 Posts	The first numbers that spring to mind are 4 and 3. Giving the product 12 and the sum 7. 12 can be expressed as 43 or 62 so Fermat could not define them. 7 can be expressed as 4+3 or 5+2 likewise for Mersenne. The statement did say the numbers where greater than 1 and other numbers would not give such a simple addition. I have not thought this completely through but it seems to give a quick solution. Regards Patrick Last fiddled with by Patrick123 on 2007-02-21 at 13:07 Reason: clarified that the numbers were > 1

2007-02-21, 13:17	#5
akruppa "Nancy" Aug 2002 Alexandria 100110100011₂ Posts	But Fermat said he knew that Mersenne knows that Fermat can't figure out the numbers. If Fermat had been given the product 12, he would have known that Mersenne must have been given the sum 7 or 8. With the sum 7, Mersenne could not have known for certain that Fermat was stuck, as it might have admitted a=2, b=5, which Fermat could have solved... Nice little mind twister. Alex Last fiddled with by akruppa on 2007-02-21 at 13:18

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2007-02-21, 09:51	#2
davieddy "Lucan" Dec 2006 England 194A₁₆ Posts	I assume that only one pair of numbers have a product which can be formed in more than one way and the sum of the two numbers is the same for more than one case. Does this make sense? I'll try it in Basic. David

2007-02-21, 13:25	#6
akruppa "Nancy" Aug 2002 Alexandria 2,467 Posts	So I guess the sum a+b was an odd number, as any even number can (conjecturally) be written as the sum of two primes, and a+b-2 must not be prime. Alex

2007-02-21, 13:30	#7
Patrick123 Jan 2006 JHB, South Africa 157 Posts	Ah! I see what you mean. I thought it was too easy. Regards Patrick

2007-02-21, 14:49	#8
XYYXF Jan 2005 Minsk, Belarus 2⁴×5² Posts	akruppa is on the right way :)

2007-02-21, 22:09	#9
Patrick123 Jan 2006 JHB, South Africa 157 Posts	The two numbers are 9 and 21. Product 189 - Sum 30. Taking Alex's lead, I used Excel and filtered out: All Evens. All primes. All products of two primes. All numbers where n-2 was prime. I then factorized the remaining numbers and created a permutation of all the ways you can sum the numbers. There where 2 products that had the same sum with one being a cube. This is all Fermat needed to know to solve this. Regards Patrick

2007-02-22, 23:31	#10
XYYXF Jan 2005 Minsk, Belarus 2⁴·5² Posts	The answer (9, 21) is wrong :)

2007-02-23, 00:11	#11
rogue "Mark" Apr 2003 Between here and the 2²×7×227 Posts	I saw this puzzle on the net and the following line is included: Both are unequal to 1 and the sum of them is less than 100. This is different than XYYXF's statement. I don't know if this is supposed to be part of this puzzle or not.